Question

The value of $\underset{x\to 0}{lim}\frac{1}{x^3}{\int }_0^x\frac{tlog(1+t)}{t^4+4}dt$ is

• A. 0
• B. $\frac{1}{12}$
• C. $\frac{1}{24}$
• D. $\frac{1}{64}$

Answer 1

Answer: (B)

$\frac{1}{12}$

We have
$\underset{x\to 0}{lim}\frac{1}{x^3}{\int }_0^x\frac{t\ln (1+t)}{t^4+4}dt\underset{x\to 0}{lim}\frac{x\ln (1+x)}{(x^4+4){3x}^2}$
$=\underset{x\to 0}{lim}\frac{1}{3}\frac{\ln (1+x)}{x(x^4+4)}=\frac{1}{12}.$